# Why are these Pascal's triangle numbers in this simple polynomial series?

I was looking at values of $$x^2 + 2*x + 1$$. Let's call it $$f$$. I'm just curious about it from a recreational point of view. It is as simple a polynomial as I could imagine, and I was looking at some of its values for small $$x$$, when I noticed something curious.

Both $$121$$ and $$14641$$ are included in the values of this series, in the 10th and 120th places respectively.

Is this just incidental? Is there any number theory connection that might explain why the distribution of digits in this series include these 'special-looking' values? Presumably numbers do things like this all the time, but it did seem sort of strange to me.

I guess I could mention that I do understand that the specific form of the number as a series of digits is itself an artifact of the base we're choosing to represent the value. I guess another way to ask the question is whether number theory has anything to say about the structure of this function and why it might contain these sorts of values, or whether it's 'entirely' an artifact.

Here are the elements of that series ($$f(x)=(x+1)^2$$) up to f(120) that I've calculated (starting with $$f(1)=1^2+2*1+1=4$$):

4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2704, 2809, 2916, 3025, 3136, 3249, 3364, 3481, 3600, 3721, 3844, 3969, 4096, 4225, 4356, 4489, 4624, 4761, 4900, 5041, 5184, 5329, 5476, 5625, 5776, 5929, 6084, 6241, 6400, 6561, 6724, 6889, 7056, 7225, 7396, 7569, 7744, 7921, 8100, 8281, 8464, 8649, 8836, 9025, 9216, 9409, 9604, 9801, 10000, 10201, 10404, 10609, 10816, 11025, 11236, 11449, 11664, 11881, 12100, 12321, 12544, 12769, 12996, 13225, 13456, 13689, 13924, 14161, 14400, 14641


Further research indicates that palindromic values are common in the squares generally:

["1", "4", "9", "121", "484", "676", "10201", "12321", "14641", "40804", "44944", "69696", "94249",
"698896", "1002001", "1234321", "4008004", "5221225", "6948496", "100020001"]


And furthermore I have found 'triangular palindromes' in $$f(x)=x^3$$ (which contains $$1331$$) and $$f(x)=x^4$$ (where I found $$14641$$).

Just for some insight on my state of mind around this: I'm increasingly feeling like I'm missing some clear connection, but also I'm still very surprised that there's anything like this in the decimal representation of these series. Why are the binominal coefficients showing up so clearly over and over again in the squares, cubes and so on? Do these 'Pascal palindromes' show up in any base for any exponent? What kind of research would shed light on these? (Part of me is still resisting ascribing any significance to the digits on principle, since questions about digits are generally base-dependent, right?)

• Note $x^2+2x+1=(x+1)^2$ and $(10+1)^2=\binom20 10^2+\binom21 10+\binom22 1$ Jan 27, 2020 at 23:11
• And you didn’t notice the pairs $(144,441)$ and $(169,961)$ on your list? But I’m sure that you noticed that your numbers are all the squares, save $1$. Jan 27, 2020 at 23:38

Hint: You should look at the binomial expansion: $$(1+x)^n=\sum\limits_{k=0}^n \binom{n}{k}1^k\cdot x^{n-k}$$. For $$n=2$$ you obtain $$(1+x)^2=\binom{2}{0}\cdot 1^0\cdot x^{2-0}+\binom{2}{1}\cdot 1^1\cdot x^{2-1}+\binom{2}{2}\cdot 1^2\cdot x^{2-2}=x^2+2x+1$$
$$n=3:(1+x)^3=$$
$$\binom{3}{0}\cdot 1^0\cdot x^{3-0}+\binom{3}{1}\cdot 1^1\cdot x^{3-1}+\binom{3}{2}\cdot 1^2\cdot x^{3-2}+\binom{3}{3}\cdot 1^3\cdot x^{3-3}=x^3+3x^2+3x+1$$
$$\ldots$$
• Okay, I think I'm starting to see it. These values are like the bases' own "reflection" of the binomial coefficients -- as J.W. Tanner pointed out, if I understand what he's getting at. I guess we'd expect to see similar patterns in other bases then, too, if I'm following? (This explains some of the structure of the 'spacier' square palindromes I suppose! $100020001$ and the like) Jan 28, 2020 at 21:32